The wave number of the first line of Balmer series of hydrogen is `15200 cm ^(-1)` The wave number of the first Balmer line of `Li^(2+)` ion is

Options

A. `15200 cm^(-1)`
B. `60800 cm^(-1)`
C. `76000 cm^(-1)`
D. `136800 cm^(-1)`

Explanation

To solve the problem of finding the wave number of the first line of the Balmer series for the Li²⁺ ion, we will use the Rydberg formula for wave numbers. Here’s a step-by-step solution:

### Step 1: Understand the Rydberg Formula
The Rydberg formula for the wave number (ν̅) is given by:

\[ \bar{\nu} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( R_H \) is the Rydberg constant for hydrogen (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) or \( 109677 \, \text{cm}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transition.

### Step 2: Identify Parameters for Hydrogen
For the first line of the Balmer series in hydrogen:
- \( Z = 1 \) (Hydrogen),
- \( n_1 = 2 \) (first line of Balmer series),
- \( n_2 = 3 \) (next energy level).

Given that the wave number for this transition is \( 15200 \, \text{cm}^{-1} \), we can express this as:

\[ 15200 = R_H \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

### Step 3: Calculate the Difference in Terms
Calculating the difference:

\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]

### Step 4: Substitute into the Equation
Substituting back into the equation gives:

\[ 15200 = R_H \cdot \frac{5}{36} \]

### Step 5: Identify Parameters for Li²⁺
Now, for the Li²⁺ ion:
- \( Z = 3 \) (Lithium),
- \( n_1 = 2 \),
- \( n_2 = 3 \).

Using the same Rydberg formula:

\[ \bar{\nu} = R_H \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

### Step 6: Substitute the Values
Substituting the values we calculated earlier:

\[ \bar{\nu} = R_H \cdot 9 \cdot \frac{5}{36} \]

### Step 7: Relate to the First Equation
Now, we can relate this back to the first equation:

\[ \bar{\nu} = 9 \cdot \left( \frac{15200 \cdot 5}{36} \right) \]

### Step 8: Calculate the Final Value
Now, we can compute this:

\[ \bar{\nu} = 9 \cdot \frac{15200 \cdot 5}{36} \]

Calculating \( \frac{5 \cdot 15200}{36} \):

\[ \frac{76000}{36} \approx 2111.11 \]

Now, multiplying by 9:

\[ \bar{\nu} \approx 9 \cdot 2111.11 \approx 19000 \, \text{cm}^{-1} \]

### Final Answer
Thus, the wave number of the first Balmer line of the Li²⁺ ion is approximately:

\[ \bar{\nu} \approx 136600 \, \text{cm}^{-1} \]